Example 3621

Using the mean values for the parameters in Table 36.8, devise a deterministic preventive maintenance plan for a bridge with respect to shear that will extend its useful life by 10 years. Assume that the minimum acceptable reliability index is btarget = 3.0. If each application of preventive maintenance costs $3,000, what is the present value preventive maintenance cost using a discount rate of 4%?

Answer

For these values, the current life of the bridge structure is b - b target 8 . 5 - 3 . 0

If the bridge life is to be extended 10 years, then a preventive maintenance program will be designed such that tRP = 79 years. The preventive maintenance cycle, tP, lasts 15 years and the change in reliability index during that time is

Dbreqd = 10a = 10(0 .1025) = 1.025 The number of maintenance applications (n) to achieve this is

Dbreqd 1 . 025

Db 0 975

Therefore, two preventive maintenance actions must be applied (n = 2). One cost-effective strategy is to schedule the maintenance as late in the life of the structure as possible. One approach is to gain as much as possible from the positive effect of the final preventive maintenance action which lasts tPD = 10 years. That would suggest applying the second maintenance action at year 69 and the first maintenance action 15 years earlier, at year 54. Figure 36.16 illustrates this strategy, which combines scheduled maintenance late in the life of the structure and maintains the reliability at a high level. The present value of preventive maintenance cost (CPM), discounted at year 0 (i.e., time of construction), is

The plan should also be checked with respect to the moment failure mode.

In many real world examples, it would not make sense to move preventive maintenance until the end of the life of the structure, but the strategy makes sense in the limited context of the problem. If failure costs were included in the analysis, there may be advantages to keeping the lifetime reliability of the structure higher and thus maintaining the structure sooner.

36.8.2 Repairs

The strategy for making repairs is similar to that for preventive maintenance as different options and their effects must be evaluated. The costs of some structural repairs are functions of the level of deterioration. Automobile body work is an example, where the amount of filling, sanding, and addition of extra material is proportional to the amount of deterioration. The costs of other structural repairs are independent of the level of structural deterioration, as, for example, adding a reinforcing plate or replacing a member, where the cost is the same regardless of the degree of deterioration. Choosing the proper mix of repairs and preventive maintenance and then timing their application in an efficient manner can be a difficult challenge.

Original Extended life life

FIGURE 36.16 Preventive maintenance plan to extend the life of a bridge deck by 10 years. EXAMPLE 36.22

Original Extended life life

FIGURE 36.16 Preventive maintenance plan to extend the life of a bridge deck by 10 years. EXAMPLE 36.22

How can one determine the best combination of repair and maintenance options for a given structure or category of structures?

Answer

Kong and Frangopol (2003b) looked at eight different maintenance scenarios and used a program LCADS (Life-Cycle Analysis of Deteriorating Structures) to predict the reliability profiles for an individual bridge associated with each maintenance strategy. The eight scenarios shown in Figure 36.17 include both preventive and essential maintenance. Sample combinations include no maintenance (E0), three applications of essential maintenance (E3), and two applications of essential maintenance with periodic preventive maintenance (E2,P). Figure 36.18 shows the mean system reliability index

FIGURE 36.17 Eight different maintenance scenarios for a steel/concrete bridge involving various combinations of preventive and essential maintenance. (Kong and Frangopol 2003b. Reprinted with permission from the American Society of Civil Engineers.)

FIGURE 36.18 Mean reliability profile associated with different maintenance scenarios. (Kong and Frangopol 2003b. Reprinted with permission from the American Society of Civil Engineers.)

profiles associated with the different scenarios over time. Data from groups of existing steel and concrete bridges were the basis for determining deterioration rates, repair effects, and associated costs.

36.8.3 Expected Life Cycle Maintenance Cost

Frangopol and Kong (2001) and Kong and Frangopol (2003a) provided a computational procedure for the evaluation of the expected life cycle maintenance cost of deteriorating structures by considering the uncertainties associated with the application of subsequent maintenance actions. The methodology can be used to determine the expected number of maintenance interventions on a deteriorating structure, or a group of deteriorating structures, during a specified time horizon and the associated expected maintenance costs. The method is suitable for application to both new and existing civil infrastructures under various maintenance strategies. The ultimate objective is to evaluate the costs of alternative maintenance strategies and determine the optimum maintenance regime over a specified time horizon. In its present format, the first line of application of the method is for highway bridges. However, the method can be used for any structure, or group of structures, requiring maintenance in the foreseeable future. The proposed method can be programmed and incorporated into an existing software package for life cycle costing of civil infrastructures.

36.9 Discount Rate

As shown in several earlier examples, the choice of repair and the timing of both inspection and repair can be highly dependent on the discount rate of money. With a higher discount rate, expenditures and investments made later in the life of a structure become more attractive. There is considerable debate as to what the correct discount rate should be. Some governments mandate a rate (such as 6%) that must be used on all public civil infrastructure projects, which at least ensures consistency between these